The proof is incorrect. In the fist part of the proof, the $\delta$ you picked does not guarantee that for all measurable sets $A$, $\mu(A) < \delta$ implies $|\int_A f_n\, d\mu| < \epsilon$. The second part of the proof is almost correct, but the condition $\lim\limits_{M\to \infty} \mu(E_M) = 0$ follows from the assumption $\sup_n \int |f_n|\, d\mu < \infty$, not that $\mu$ is finite. Indeed, if $\alpha := \sup_n \int |f_n|\, d\mu < \infty$, then $\mu(E_M) \le \frac{\alpha}{M}$ for all $M > 0$; as a consequence, $\lim\limits_{M\to \infty} \mu(E_M) = 0$. If you make this fix, then the second part of the proof will be correct.
To prove the forward direction, fix $\epsilon > 0$ and choose $M > 0$ such that
$$\sup_n \int_{E_M} |f_n| \, d\mu < \frac{\epsilon}{2}.$$
Then
$$\int |f_n|\, d\mu = \int_{E_M} |f_n|\, d\mu + \int_{E_M^c} |f_n|\, d\mu < \frac{\epsilon}{2} + M\mu(X) < \infty$$
for all $n\in N$.
Set $\delta = \frac{\epsilon}{2M}$. For all measurable sets $A$, $\mu(A) < \delta$ implies
$$\left|\int_A f_n\, d\mu\right| \le \int_A |f_n|\, d\mu = \int_{A\cap E_M} |f_n|\, d\mu + \int_{A\cap E_M^c} |f_n|\, d\mu < \frac{\epsilon}{2} + M\mu(A) < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon.$$
Since $\epsilon$ was arbitrary, $\{f_n\}$ is uniformly absolutely continuous.
Since $f_n$ converges uniformly on $\Omega$ to $f$, for $\forall \varepsilon>0$, there is $N>0$ which is independent of $x\in\Omega$ such that
$$ |f_n(x)-f(x)<\frac{\varepsilon}{\mu(\Omega)}, \forall n\ge N, x\in\Omega. $$
So for above $\varepsilon>0$ and $N>0$, when $n\ge N$,
$$\bigg|\int_\Omega (f_n - f)\, d\mu\bigg|\le\int_\Omega |f_n-f|\ d\mu\le\int_{\Omega}\frac{\varepsilon}{\mu(\Omega)}d\mu=\varepsilon $$
which implies
$$ \lim_{ n \to \infty}\int_\Omega f_n \, d\mu=\int_\Omega f_n \, d\mu. $$
Best Answer
Because $\mu (\Omega) < \infty$ the constant function $1_{\Omega}$ on $\Omega$ is integrable ($\int 1_{\Omega}d\mu = \mu(\Omega)$) and thereby also $f_n + \epsilon$ and $f_n-\epsilon$ by linearity of the integral. As the $f_n$ converge uniformly, for all $\epsilon>0$ there exists $n_0$ such that for all $n>n_0$ $|f_n-f|<\epsilon$ and so $f_n-\epsilon<f<f_n+\epsilon$, so $f$ is bounded by an integrable function and therefor integrable (This follows from the dominance convergence theorem and the fact that $\mu(\Omega)<\infty$, so every integrable function is absolutely integrable).
But now for the integral we get by monotony that:
$\int f_nd\mu - \epsilon*\mu(\Omega) =\int (f_n-\epsilon) d\mu < \int fd\mu<\int (f_n+\epsilon) d\mu<\int f_nd\mu + \epsilon*\mu(\Omega)$ for $n>n_0$
As $\epsilon>0$ is arbitrary this gives for all $\delta>0$ an $n_0$ such that for all $n>n_0$ $|\int(f_n-f)d\mu|<\delta$ if we choose $\epsilon=\delta/\mu(\Omega)$, so the integral converges.